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Let $y=f(x)$ be an infinitely differentiable function on real numbers such that $f(0)$ is not equal to 0 and $d^n(y)∕dx^n$ not equal to zero at $x=0$ for $n=1,2,3,4$. If $$\lim_{x→0} {f(4x) + af(3x) + bf(2x) + cf(x) + df(0)\over x^4} $$ exists then find the value of $25a + 50b +100c + 500d$.

The answer is :

$300$

but I can't solve it.

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  • $\begingroup$ We don't really need $f$ to be infinitely differentiable, just existence of fourth derivative at $x = 0$ is fine. But its OK if the question is giving more data than necessary. $\endgroup$ – Paramanand Singh May 13 '16 at 7:26
  • $\begingroup$ It would help if you can upvote my answer (which you have accepted) because although it is perfectly correct it is suffering from a downvote by some user. $\endgroup$ – Paramanand Singh May 15 '16 at 7:55
  • $\begingroup$ I am new here so I am short of reputation points. Once I cross them I would do it. $\endgroup$ – Harry Karwasra May 15 '16 at 9:53
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Well the right tool to use here is Taylor's series and not L'Hospital's Rule. Note that we don't need anything more than the existence of $f^{(4)}(0)$ and the fact that all of $f(0), f'(0), f''(0), f'''(0), f^{(4)}(0)$ are non-zero.

We have via Taylor's series $$f(x) = f(0) + xf'(0) + \frac{x^{2}}{2}f''(0) + \frac{x^{3}}{6}f'''(0) + \frac{x^{4}}{24}f^{(4)}(0) + o(x^{4})\tag{1}$$ Replacing $x$ by $2x, 3x, 4x$ respectively we get \begin{align} f(2x) &= f(0) + 2xf'(0) + 2x^{2}f''(0) + \frac{4x^{3}}{3}f'''(0) + \frac{2x^{4}}{3}f^{(4)}(0) + o(x^{4})\tag{2}\\ f(3x) &= f(0) + 3xf'(0) + \frac{9x^{2}}{2}f''(0) + \frac{9x^{3}}{2}f'''(0) + \frac{27x^{4}}{8}f^{(4)}(0) + o(x^{4})\tag{3}\\ f(4x) &= f(0) + 4xf'(0) + 8x^{2}f''(0) + \frac{32x^{3}}{3}f'''(0) + \frac{32x^{4}}{3}f^{(4)}(0) + o(x^{4})\tag{4} \end{align} Hence we have the numerator of the limit expression given by \begin{align} Nr &= f(4x) + af(3x) + bf(2x) + cf(x) + df(0)\notag\\ &= f(0)(1 + a + b + c + d) \notag\\ &\,\,\,\,+ xf'(0)\left(4 + 3a + 2b + c\right)\notag\\ &\,\,\,\,+x^{2}f''(0)\left(8 + \frac{9a}{2} + 2b + \frac{c}{2}\right)\notag\\ &\,\,\,\,+x^{3}f'''(0)\left(\frac{32}{3} + \frac{9a}{2} + \frac{4b}{3} + \frac{c}{6}\right)\notag\\ &\,\,\,\,+x^{4}f^{(4)}(0)\left(\frac{32}{3} + \frac{27a}{8} + \frac{2b}{3} + \frac{c}{24}\right) + o(x^{4})\notag \end{align} From the above it is clear that the limit in question exists only when the following equations are satisfied simultaneously: \begin{align} a + b + c + d &= -1\tag{5a}\\ 3a + 2b + c &= -4\tag{5b}\\ 9a + 4b + c &= -16\tag{5c}\\ 27a + 8b + c &= -64\tag{5d} \end{align} Solving these equations we get $$a = -4, b = 6, c = -4, d = 1$$ and thus the value of $$25a + 50b +100c + 500d$$ is $300$.


Update: Note that the problem can also be solved using L'Hospital's Rule but it needs more careful analysis. I wished to highlight this point in comments to another answer of this question but it was difficult to give all the details in a comment.

Since the limit $$L = \lim_{x \to 0}\frac{f(4x) + af(3x) + bf(2x) + cf(x) + df(0)}{x^{4}}\tag{6}$$ exists it is clear that as $x \to 0$ the numerator of the above fraction tends to $0$. Hence $f(0)(1 + a + b + c + d) = 0$ which implies that $$a + b + c + d = -1\tag{7}$$ Next we know that the fraction in the limit $(6)$ is of the form $0/0$ and hence there is a chance that L'Hospital Rule will work. On applying LHR we get the fraction $$\frac{4f'(4x) + 3af'(3x) + 2bf'(2x) + cf'(x)}{4x^{3}}\tag{8}$$ The numerator tends to $f'(0)(4 + 3a + 2b + c)$ and if $(4 + 3a + 2b + c) \neq 0$ then the above fraction tends to infinity as $x \to 0$. This would imply that the original limit $(6)$ will also be infinite. Thus we must have $$3a + 2b + c = -4\tag{9}$$ This means that the fraction in $(8)$ is also of the form $0/0$ (but we don't know if its limit exists or not). Hence there is a chance that further application of LHR might work. Doing this we get another fraction $$\frac{16f''(4x) + 9af''(3x) + 4bf''(2x) + cf''(x)}{12x^{2}}\tag{10}$$ The numerator tends to $f''(0)(16 + 9a + 4b + c)$ and if this is non-zero then the ratio in $(10)$ would tend to infinity and hence by LHR the ratio in $(8)$ would also tend to infinity. This would imply that the original limit in $(6)$ is also infinite which is not the case. Hence we must have $$9a + 4b + c = -16\tag{11}$$ In similar manner we can show that $$27a + 8b + c = -64\tag{12}$$ From $(7), (9), (11), (12)$ we can find values of $a, b, c, d$ and get the same answer $300$.

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  • $\begingroup$ Would the downvoter care to comment?? $\endgroup$ – Paramanand Singh May 14 '16 at 4:25
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Since the limit exists, denote it with $L$.

$$L = \lim_{x→0} {f(4x) + af(3x) + bf(2x) + cf(x) + df(0)\over x^4}$$

If we attempt to take the limit, we naively may think

$$L = {f(0)(1 + a + b + c + d)\over 0^4}$$

If the numerator is nonzero, then the expression diverges. Yet, $L$ exists. Therefore we expect the indeterminate form $\frac00$. The numerator must equal zero. Specifically, since $f(0)\ne 0$,

$$1 + a + b + c + d=0$$

Since the limit gives an indeterminate form, we employ L'Hopital as well:

$$L = \lim_{x→0} {4f'(4x) + 3af'(3x) + 2bf'(2x) + cf'(x) \over 4x^3}$$

This time, with a bit of work we can conclude

$$4+3a+2b+c = 0$$

Can you take it from here?

EDIT:

If you're concerned about the application of L'Hopital, convince yourself that the above steps are justified recursively, as is the case with many applications of L'Hopital's rule, upon reaching the finite limit $L$. This post is clearly not a thorough proof, but a guide to the problem at hand.

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    $\begingroup$ The limit doesn't diverge, the expression does. $\endgroup$ – zhw. May 13 '16 at 3:51
  • $\begingroup$ Thanks. Will edit. That is more accurate phrasing. $\endgroup$ – zahbaz May 13 '16 at 3:53
  • $\begingroup$ this does not work as we don't know if the limit after applying L'Hospital's Rule exist or not. The proper mechanism is to use Taylor's series. There is a way to salvage this with more careful analysis of each limit obtained after applying L'Hospital's Rule, but a naive application of the rule as you have done is not correct. $\endgroup$ – Paramanand Singh May 13 '16 at 7:20
  • $\begingroup$ @ParamanandSingh Why wouldn't it? If $\lim_{x\rightarrow0}f(x)\ne0$, then the original limit must have reached an indeterminate form if $L$ is to exist. The conditions for L'Hopital can be inferred, and we can apply the rule successively until the denominator reaches 1. What alternative outcome could occur? $\endgroup$ – zahbaz May 13 '16 at 8:56
  • $\begingroup$ No! L'Hospital rule uses the existence of $\lim f'/g'$ to infer existence of $\lim f/g$. Here you know that $\lim f/g$ exists and you then also assume that $\lim f'/g'$ also exists. This is not the way L'Hospital works. $\endgroup$ – Paramanand Singh May 13 '16 at 9:00

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